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bogdanovich [222]
3 years ago
5

PLEASE HELP!!

Physics
1 answer:
defon3 years ago
7 0
The right answer is A
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The common uses of radioactive elements in health care.
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A sled and its rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow. The snow exerts a kinetic frictional f
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Answer:

The acceleration of the sled is "0.49 m/s²". A further explanation is given below.

Explanation:

The given values are,

Speed,

V = 4.0 m/s

Coefficient of kinetic friction,

μ = 0.05

As we know,

⇒  F \mu=\mu mg

and,

⇒  a = \mu g

On substituting the given values, we get

⇒     =0.0 5\times 9.8

⇒     =0.49 \ m/s^2

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3 years ago
Maria drove 100 miles in 2 hrs. what was marias speed?​
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50 miles per hour was Maria’s speed
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a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
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